# Requesting help with centripetal acceleration homework

## Homework Statement

A hockey puck of mass m = 80 g is attached to a string that passes through a hole in the center of a table, as shown in the figure below. The hockey puck moves in a circle of radius r = 1.10 m. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M = 1.00 kg. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?

M puck = 0.08kg
M block = 1kg
Gravity = 9.8
r = 1.10

F=ma
Ac = (v^2/r)

## The Attempt at a Solution

Attached in photo, why the gravity force of puck not being included is confusing me alot, shouldn't all objects be affected by gravity?

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Attached in photo, why the gravity force of puck not being included is confusing me alot, shouldn't all objects be affected by gravity?
Do you feel gravitational force on the buck has any influence on the tension of the string? Are they not acting perpendicular to each other over the buck? Moreover, the table surface is smooth, right?

You'll need someone else to verify if this is correct or not as my physics know-how is a bit dated these days. However, I believe gravity comes into play in the form of friction against the surface. Since we're assuming the tabletop is perfectly smooth, gravity on the puck is no longer a factor because there is no friction. Gravity's pull in this case is counteracted by the table.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

A hockey puck of mass m = 80 g is attached to a string that passes through a hole in the center of a table, as shown in the figure below. The hockey puck moves in a circle of radius r = 1.10 m. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M = 1.00 kg. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?

M puck = 0.08kg
M block = 1kg
Gravity = 9.8
********** 9.8 what? **************
r = 1.10
********* 1.10 what ****************

F=ma
Ac = (v^2/r)

## The Attempt at a Solution

Attached in photo, why the gravity force of puck not being included is confusing me alot, shouldn't all objects be affected by gravity?

Units are important; do not state units for two of the data items and omit them for two others.

haruspex
Homework Helper
Gold Member
2020 Award
why the gravity force of puck not being included
Because the normal force from the table is equal and opposite to it, so they cancel. And as others have mentioned, there is no friction.

Thank you for all your responses, I am able to understand it now.

Also next time I will make sure to add all the units to my variables.